Problem: Baby Ehab was toying around with arrays. He has an array a of length n. He defines an array to be good if there's no way to partition it into 2 subsequences such that the sum of the elements in the first is equal to the sum of the elements in the second. Now he wants to remove the minimum number of elements in a so that it becomes a good array. Can you help him?

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly zero or all) elements. A partitioning of an array is a way to divide it into 2 subsequences such that every element belongs to exactly one subsequence, so you must use all the elements, and you can't share any elements.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 2000) — the elements of the array a.

Output

The first line should contain the minimum number of elements you need to remove.

The second line should contain the indices of the elements you're removing, separated by spaces.

We can show that an answer always exists. If there are multiple solutions, you can print any.

Examples

Input


4
6 3 9 12


Output


1
2

Input


2
1 2


Output


0

Note

In the first example, you can partition the array into [6,9] and [3,12], so you must remove at least 1 element. Removing 3 is sufficient.

In the second example, the array is already good, so you don't need to remove any elements.

Solution (in CPP):

Solution:
/*...................bismillahir rahmanir rahim.....................*/
/*
  author :
    Muhammad Najmul hasan Nayeem
    Sust Math 17
*/
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/trie_policy.hpp>
// using namespace __gnu_pbds;
 
using namespace std;

#define m_p make_pair
#define PB push_back
#define eb emplace_back
#define FF first
#define SS second
#define MX5 100005
#define MX6 1000006
#define MOD 1000000007
#define in()({int a;scanf("%d",&a);a;})
#define SI(a) scanf("%d",&a)
#define SII(a,b) scanf("%d%d",&a,&b)
#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define SC()({char a;scanf("%c",&a);a;})
#define SL()({long long a;scanf("%lld",&a);a;})
#define SD()({double a;scanf("%f",&a);a;})
#define nl '\n'
#define For(_,n) for(int i=_;i<=n;i++)
#define once "once upon a time\n"
#define FI freopen("input.txt","r+",stdin);
#define FO freopen("output.txt","w+",stdout);
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<string,string> PSS;
// typedef tree<PII,null_type,less<PII>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

template <class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {
    return out << "(" << a.FF << ", " << a.SS << ")";
}
 
template <class A> ostream& operator << (ostream& out, const vector<A> &v) {
    out << "[";
    for(int i=0;i<v.size();i++) {
        if(i) out << ", ";
        out << v[i];
    }
    return out << "]";
}
 
#define chk(a) cout<<(#a)<<" = "<<a<<endl;
#define chka(a,_) cout<<(#a)<<"["<<_<<"] = "<<a[_]<<"\n";
// #define debug(args...)                           \
//     {                                            \
//         string _s = #args;                       \
//         replace(_s.begin(), _s.end(), ',', ' '); \
//         stringstream _ss(_s);                    \
//         istream_iterator<string> _it(_ss);       \
//         deb(_it, args);                          \
//     }
// void deb(istream_iterator<string> it)
// {
// }
// template <typename T, typename... Args>
// void deb(istream_iterator<string> it, T a, Args... args)
// {
//     cout << *it << " = " << a << endl;
//     deb(++it, args...);
// }
// int Set(int n,int pos){ return n | (1<<pos); }
// int Reset(int n,int pos){ return n=n & ~(1<<pos); }
// bool Check(int n,int pos){ return (bool)(n & (1<<pos)); }
struct GCD_type { LL x, y, d; };
GCD_type ex_GCD(LL a, LL b)
{
    if (b == 0) return {1, 0, a};
    GCD_type pom = ex_GCD(b, a % b);
    return {pom.y, pom.x - a / b * pom.y, pom.d};
}
/*............start.............*/
int n;
int a[101];
int t;
int dp[101][200005];
int nh(int i,int sum)
{
    if(i==n) return (sum==t);
    int &ret=dp[i][sum];
    if(ret!=-1) return ret; 
    ret=0;
    ret|=(bool)nh(i+1,sum+a[i]);
    ret|=(bool)nh(i+1,sum);
    return ret;
}
void solve()
{
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    // sort(a,a+n);
    int sum=0;
    for(int i=0;i<n;i++)
    {
        sum+=a[i];
    }
    if(sum&1)
    {
        cout<<0<<nl;
        return;
    }
    t=sum/2;
    memset(dp,-1,sizeof(dp));
    if(!nh(0,0))
    {
        cout<<0<<nl;
        return;
    }
    for(int i=0;i<n;i++)
    {
        if(a[i]&1)
        {
            cout<<1<<nl<<i+1<<nl;
            return;
        }
    }
    while(1)
    {
        for(int i=0;i<n;i++)
        {
            if(a[i]&1)
            {
                cout<<1<<nl<<i+1<<nl;
                return;
            }
            a[i]/=2;
        }
    }

}

int main()
{
    // clock_t time_start=clock();
    ios_base::sync_with_stdio(false);//cin.tie(NULL);cout.tie(NULL);
    int q=1;
    // cin>>q;
    // SI(q);
    for(int kase=1;kase<=q;kase++)
    {
        // cout<<"Case #"<<kase<<": ";
        // printf("Case %d: ",kase );
        solve();
    }
    // clock_t time_end=clock();
    // cerr<<float((time_end-time_start)/CLOCKS_PER_SEC);
    return 0;
}